TITLE: Why $S$ cannot be homeomorphic to the $1$-sphere of $\ell^2$?
QUESTION [8 upvotes]: Consider the $\ell^2$ complex Hilbert space.
Let $m\in \mathbb{N}^*$ be a fixed number, and set 
$$
S=\left\{ x=(x_n)_n\subset \ell^2\ :\ \sum_{n=1}^m \frac{|x_n|^2}{n^2}=1\right\}.$$

I want to show that $S$ is not homeomorphic to 
  $$
S(0,1)=\left\{ x=(x_n)_n\subset \ell^2\ :\ \sum_{n=1}^\infty |x_n|^2=1\right\}.$$

REPLY [8 votes]: $S$ contains the ellipsoid $\{x\in \mathbb C^m: \sum \frac{|x_n|^2}{n^2}=1\}$ (which is homeomorphic to $S^{2m-1}$) as a strong deformation retract via
$A_t(x_1,\dots,x_m,x_{m+1},\dots) = (x_1,x_2,\dots,x_m,(1-t)x_{m+1},(1-t)x_{m+2},\dots)$ for $t\in[0,1]$, thus the homotopy group $\pi_{2m-1}(S) = \mathbb Z$.
$S(0,1)$, however, has $\pi_{2m-1}(S(0,1)) = 0$ since it is contractible: shift first all coordinates one step to the right and then contract in the stereographic chart opposite $(1,0,0,\dots)$.
Or use the contraction described on page 513 of here.